Question: $B$ and $C$ trisect $\overline{AD}$ and $M$ is the midpoint of $\overline{AD}$. $MC = 8$. How many units are in the length of $\overline{AD}$?
Solution: Since $MC = 8$ and $M$ is the midpoint of $\overline{BC}$, we have $MB=MC = 8$, so $BC=8+8=16$.  Since $B$ and $C$ trisect $\overline{AD}$, we have $AB = CD = BC = 16$, so $AD =16+16+16=\boxed{48}$.